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Math Help - Analytic Geometry Q10

  1. #1
    Member looi76's Avatar
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    Analytic Geometry Q10

    Question:
    Find the equation of the bisector of the pair of acute angles formed by the lines L_1 and L_2 given by 4x-2y-10=0 and x-2y-1=0, respectively.

    Attempt:

    L_1 \rightarrow 4x - 2y - 15 = 0
    A = 4, \ B = -2, \ C=-15

    d_1 = \frac{Ax+By+C}{\pm\sqrt{A^2+B^2}} = \frac{4x-2y-15}{-\sqrt{4^2+2^2}} = \frac{4x-2y-15}{-2\sqrt{5}}

    L_2 \rightarrow x - 2y - 1
    A = 1, \ B = -2, \ C=-1

    d_2 = \frac{Ax+By+C}{\pm\sqrt{A^2+B^2}} = \frac{x-2y-1}{-\sqrt{1^2+2^2}} = \frac{x-2y-1}{-\sqrt{5}}

    Now, I need to simplify the two distance equations to get the equation of the bisector...

    \frac{4x-2y-15}{-2\sqrt{5}} = \frac{x-2y-1}{-\sqrt{5}}

    Here is the problem! I don't know how to simplify with the square roots around , Need Help!
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  2. #2
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    Quote Originally Posted by looi76 View Post
    [FONT="Georgia"][SIZE="3"]Question:
    Find the equation of the bisector of the pair of acute angles formed by the lines L_1 and L_2 given by 4x-2y-10=0 and x-2y-1=0, respectively.

    ...
    1. Both equations are given in normal form. The normal vectors are :

    \overrightarrow{n_1} = (4, -2) . Therefore the direction vector of the first line is \vec u = (2, 4) with |\overrightarrow{u}| = \sqrt{20}

    and

    \overrightarrow{n_2} = (1, -2) . Therefore the direction vector of the second line is \vec v = (2,1) with |\overrightarrow{v}| = \sqrt{5}

    2. The point of intersection is P(3,1) . Therefore the angle bisector has to pass through this point too.

    3. The direction of the angle bisector is the sum of the unit vectors of the two direction vectors:

    \vec b = \dfrac1{\sqrt{20}} (1,4) + \dfrac1{\sqrt{5}} (2,1) = \dfrac1{\sqrt{20}} (2,4) + \dfrac1{\sqrt{20}} (4,2) = \dfrac1{\sqrt{20}} (6, 6)

    Remark: \dfrac1{\sqrt{20}} = \dfrac1{\sqrt{5} } \cdot \dfrac1{\sqrt{4}}

    The second angle bisector is perpendicular to the first one thus it has the direction \overrightarrow{b_{\perp}} = \dfrac1{\sqrt{20}} (6, -6)

    4. You can simplify those direction vectors to \vec b = (1,1) and \overrightarrow{b_{\perp}} =  (1,-1)

    5. Now use the point direction form of the equation of a line to determine the equations of the two angle bisectors.

    EDIT: I've attached a sketch of the two lines and the two angle bisectors
    Attached Thumbnails Attached Thumbnails Analytic Geometry Q10-zweiwinkhalb.png  
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  3. #3
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    Quote Originally Posted by looi76 View Post
    Question:
    Find the equation of the bisector of the pair of acute angles formed by the lines L_1 and L_2 given by 4x-2y-10=0 and x-2y-1=0, respectively.
    Another question! Does it make any difference if the pair formed an acute angle or obtuse angle?
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  4. #4
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    Quote Originally Posted by looi76 View Post
    Another question! Does it make any difference if the pair formed an acute angle or obtuse angle?
    No because two lines form two angles: one acute and one obtuse. With one exception: If the lines are perpendicular then the two angles are equal.
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  5. #5
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    Quote Originally Posted by looi76 View Post
    ...
    Now, I need to simplify the two distance equations to get the equation of the bisector...

    \frac{4x-2y-15}{-2\sqrt{5}} = \frac{x-2y-1}{-\sqrt{5}}

    ...
    I played a little bit with your equation and much to my surprise found out that this is a very effective way to calculate the equations of the angle bisectors:

    \frac{4x-2y-15}{-2\sqrt{5}} = \frac{x-2y-1}{-\sqrt{5}} Multiply by -\sqrt{20}

    4x-2y-10 = 2(x-2y-1)~\implies~y = -x+4

    \frac{4x-2y-10}{-2\sqrt{5}} = \frac{x-2y-1}{+\sqrt{5}} Multiply by -\sqrt{20}

    4x-2y-10 = -2x+4y+2~\implies~y = x-2

    Thanks for this shortcut.
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